Problem 90 - Applied Statistics and Probability for Engineers

Question

Power meters enable cyclists to obtain power measurements nearly continuously. The meters also calculate the average power generated over a time interval. Professional riders can generate 6.6 watts per kilogram of body weight for extended periods of time. Some meters calculate a normalized power measurement to adjust for the physiological effort required when the power output changes frequently. Let the random variable \(X\) denote the power output at a measurement time and assume that \(X\) has a lognormal distribution with parameters \(\theta=5.2933\) and \(\omega^{2}=0.00995 .\) The normalized power is computed as the fourth root of the mean of \(Y=X^{4}\). Determine the following: (a) Mean and standard deviation of \(X\) (b) \(f_{Y}(y)\) (c) Mean and variance of \(Y\) (d) Fourth root of the mean of \(Y\) (e) Compare \(\left[E\left(X^{4}\right)\right]^{1 / 4}\) to \(E(X)\) and comment.

Step-by-Step Solution

Answer

(a) Use lognormal formulas for mean and standard deviation of X. (b) Determine Y's PDF using transformed lognormal parameters. (c) Calculate lognormal mean and variance of Y. (d) Compute the fourth root of mean of Y. (e) Compare results from (d) and mean of X.

Step 1: Understanding Lognormal Distribution for X

Given that the random variable \(X\) has a lognormal distribution, it means that the natural logarithm of \(X\), denoted as \(\ln(X)\), is normally distributed. The parameters for \(X\) are \(\theta = 5.2933\) (mean of \(\ln(X)\)) and \(\omega^2 = 0.00995\) (variance of \(\ln(X)\)).

Step 2: Calculate Mean and Standard Deviation of X

The mean of a lognormally distributed random variable \(X\) is given by: \[ E(X) = e^{\theta + \frac{\omega^2}{2}} \]. The standard deviation is given by: \[ \sigma_X = e^{\theta + \frac{\omega^2}{2}}\sqrt{e^{\omega^2} - 1} \].

Step 3: Deriving PDF for Y = X^4

Since \(Y = X^4\), this means that \(Y\) is also lognormally distributed because any power of a lognormally distributed variable is also lognormal. The parameters are transformed as: \(\mu_Y = 4\theta\) and \(\sigma_Y^2 = 4^2 \cdot \omega^2\). Use these to find the PDF \(f_Y(y)\).

Step 4: Calculate Mean and Variance of Y

For \(Y\), use the lognormal distribution formulas: The mean \(E(Y) = e^{4\theta + 8\cdot \omega^2}\) and the variance \(Var(Y) = (e^{8\cdot \omega^2} - 1)e^{2(4\theta + 8\cdot \omega^2)}\).

Step 5: Compute Fourth Root of Mean of Y

To find the normalized power, calculate the fourth root of the mean of \(Y = X^4\). Thus compute \(\sqrt[4]{E(Y)} = \sqrt[4]{e^{4\theta + 8 \cdot \omega^2}}\).

Step 6: Compare \([E(X^4)]^{1/4}\) to \(E(X)\)

Compare the fourth root of the mean of \(Y = X^4\), computed in the previous step, with the mean of \(X\). Observations will include noting any difference or similarity between these values and what it implies about power output.

Key Concepts

Power output analysis

Power output analysis is essential in understanding the performance and efficiency of an athlete, especially cyclists who use power meters. Power meters provide almost continuous data on power output, capturing the athlete's output in real-time. This allows for calculating average, maximum, and normalized power over different intervals of a ride. Normalized power adjusts the measured power to reflect the physiological effort required for varying power outputs, helping to analyze performance more accurately.
A cyclist's power output, modeled by a random variable, is often assumed to have a lognormal distribution. This is because even minor changes in power can impact an athlete's performance significantly, reflecting a more skewed nature of data. The data skewness towards higher power outputs makes the lognormal distribution a suitable choice for analysis because it can describe data that is positively skewed. This understanding of power output data is crucial as it influences training programs and race strategies.

Mean and standard deviation calculation

Calculating the mean and standard deviation of a lognormally distributed variable is critical as it translates the skewed data back into a more interpretable format. For a lognormal variable like the power output of a cyclist, with parameters \(\theta\) and \(\omega^2\), the mean \(E(X)\) is calculated using the formula:

• \(E(X) = e^{\theta + \frac{\omega^2}{2}}\)

The formula accounts for the back-transformation from the normal distribution on the log scale to the original scale of measurements.
The standard deviation \(\sigma_X\) provides insight into the variability of the power output:

• \(\sigma_X = e^{\theta + \frac{\omega^2}{2}}\sqrt{e^{\omega^2} - 1}\)

Calculating these statistics is essential for interpreting the average power generated and assessing consistency in power output, indicating the athlete's ability to maintain performance during various phases of cycling.

Probability density function derivation

The probability density function (PDF) provides the likelihood of the random variable, like power output, taking on specific values. When a lognormal variable \(X\) is raised to a power, in this case, \(Y = X^4\), it remains lognormally distributed. The transformation modifies the parameters accordingly:

• The new mean parameter \(\mu_Y = 4\theta\)
• The variance parameter \(\sigma_Y^2 = 4^2 \cdot \omega^2\)

Utilizing these transformed parameters, the PDF \(f_Y(y)\) can be derived and analyzed for \(Y\), giving insights into the distribution of higher power outputs and the impact of varying conditions on power. Understanding how to derive and interpret the PDF helps characterize the output's uncertainty and evaluate the probability of different performance levels, aiding in strategic decision-making.

Variance calculation

Variance measures the degree of spread or variability within a set of data. In the context of a lognormal distribution, variance is a crucial metric as it reflects how much the power output can deviate from the mean, indicating unpredictability or consistency. For the variable \(Y = X^4\), the variance is computed with the formula:

• \(Var(Y) = (e^{8 \cdot \omega^2} - 1)e^{2(4\theta + 8 \cdot \omega^2)}\)

This formula accounts for the exponential nature of a lognormal distribution and the scaling transformation caused by the exponentiation. Calculating variance allows athletes and coaches to understand the range of power outputs, preparing better for different racing scenarios and optimizing training regimens, to harness consistent performance results. Variance thus plays a vital role in guiding practice efforts by indicating stable or erratic power patterns over time.